Complementary Angles
Theorem: \({{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2},\ for\ \ all\ \ x\in \left[ -1,1 \right]\),
Proof: Let \({{\sin }^{-1}}x=\theta \). . .(1)
Where \(x\in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\),
\(\Rightarrow -\frac{\pi }{2}\le -\theta \le \frac{\pi }{2}\),
\(\Rightarrow -\frac{\pi }{2}+\frac{\pi }{2}\le -\theta +\frac{\pi }{2}\le \frac{\pi }{2}+\frac{\pi }{2}\),
\(\Rightarrow 0\le \frac{\pi }{2}-\theta \le \pi \),
\(\Rightarrow\frac{\pi }{2}-\theta \in \left[ 0,\pi \right]\),
Now sin⁻¹x = θ
x = sin θ
x = cos (π/2 – θ)
cos⁻¹x = π/2 – θ
cos⁻¹x + θ = π/2. . . (2)
from equation 1 and 2
sin⁻¹x = θ
cos⁻¹x + θ = π/2
cos⁻¹x + sin⁻¹x = π/2
Note:
- Tan⁻¹x + cot⁻¹x = π/2, for all x ϵ R.
- sec⁻¹x + cosec⁻¹x = π/2, for all x ϵ (-∞, -1] \(\cup \) [1, ∞).
Example: If sin⁻¹x = π/5, for some x ϵ (-1, 1), then find the value of cos⁻¹x.
Solution:
sin⁻¹x = π/5
cos⁻¹x + sin⁻¹x = π/2
cos⁻¹x + π/5= π/2
cos⁻¹x = π/2 – π/5
cos⁻¹x = 3 π/10